'''
可见的点的特点是两个坐标互质，利用线性筛法从2到N求每个数的欧拉函数，然后求欧拉函数总和
基于对称性，再乘以2，最后再加(1,1), (0, 1), (1, 0)三个特殊点即可
'''

def get_eular_vals(N):
    prime_vals = []
    phi_vals = [0]*(N+1)      # 每一个数的欧拉函数数值
    phi_vals[1] = 1
    flag = [True] * (N+1)     # 是否是质数标记

    # 下面是线性筛法选质数，每个数的欧拉函数数值被其最小的质因数更新
    for val in range(2, N+1):
        if flag[val]:
            prime_vals.append(val)
            phi_vals[val] = val-1

        for p_val in prime_vals:
            if val * p_val > N:
                break

            flag[val * p_val] = False

            if val % p_val == 0:
                # p_val能整除val, p_val*val 的欧拉函数相对val的欧拉函数就是多乘了一项p_val
                phi_vals[val * p_val] = phi_vals[val] * p_val
                break
            else:
                # p_val不能能整除val, p_val*val 的欧拉函数相对val的欧拉函数就是多乘了一项p_val * (1- 1/p_val)
                phi_vals[val * p_val] = phi_vals[val] * (p_val-1)

    return phi_vals

eular_vals = ans = get_eular_vals(1000)


n = int(input())
for i in range(1, n+1):
    val = int(input())
    print(i, val, sum(eular_vals[2:val+1])*2+3)